Given two horizontal planes cut by a vertical line, it is required to and a point on the upper part of the vertical line from which bodies may fall to the horizontal planes and there, having their motion deflected into a horizontal direction, will, during an interval equal to the time of fall, traverse distances which bear to each other any assigned ratio of a smaller quantity to a larger.

Let CD and BE be the horizontal planes cut by the vertical ACB, and let the ratio of the smaller quantity to the larger be that of N to FG. It is required to find in the upper part of the vertical line, AB, a point from which a body falling to the plane CD and there having its motion deflected along this plane, will traverse, during an interval equal to its time of fall a distance such that if another body, falling from this same point to the plane BE, there have its motion deflected along this plane and continued during an interval equal to its time of fall, will traverse a distance which bears to the former distance the {266} ratio of FG to N. Lay off GH equal to N, and select the point L so that FH:HG = BC:CL. Then, I say, L is the point sought. For, if we lay off CM equal to twice CL, and draw the line LM cutting the plane BE at O, then BO will be equal to twice BL. And since FH:HG = BC:CL, we have, componendo et convertendo, HG:GF = N:GF = CL:LB = CM:BO. (Condition 2/23-pr-09-schol1) It is clear that, since CM is double the distance LC, the space CM is that which a body falling from L through LC will traverse in the plane CD; and, for the same reason, since BO is twice the distance BL, it is dear that BO is the distance which a body, after fall through LB, will traverse during an interval equal to the time of its fall through LB. Q. E. F.